 Solution

ID: past-exam-of-the-numerical-analysis-course-of-the-university-of-oxford/2017/1/b/solution

x_{k}

gives

w^{'} (x_{k}) = \prod_{j \neq = k} (x_{k} - x_{j}) \neq = 0, ℓ_{k} (x) = \frac{w ( x )}{( x - x _{k} ) w ^{'} ( x _{k} )} = w (x) \frac{w _{k}}{x - x _{k}} .

p_{n} (x) = w (x) \sum_{k = 0}^{n} \frac{w _{k} f _{k}}{x - x _{k}} .

1 = \sum_{k = 0}^{n} ℓ_{k} (x) = w (x) \sum_{k = 0}^{n} \frac{w _{k}}{x - x _{k}} .

Dividing the preceding two identities proves the barycentric interpolation formula

p_{n} (x) = \frac{k = 0 \sum n w _{k} f _{k} / ( x - x _{k} )}{k = 0 \sum n w _{k} / ( x - x _{k} )} .

The denominator is exactly

1/ w (x)

, so is nonzero away from the interpolation nodes. At

x = x_{j}

x - x_{j}

and take the limit: their values tend to

w_{j} f_{j}

and

w_{j}

, respectively. The removable singularity therefore has value

f_{j}

; in numerical evaluation return the stored value directly at an interpolation node.

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